I have 108 variables with same name but with consecutive numbering. Each of this variables I want to check if there is a value bigger 1000. When yes, then it will be exchanged through NaN.
I try to solve with eval and for-loop, but it doesn’t work.
I know the function which changes a value bigger 1000 through NaN.
Function: Temperatur_1(Temperatur_1>1000)=NaN;
Dynamics variabels are: Temperatur_1, Temperatur_2, etc.
eval-function: eval(strcat(‘Temperatur_’,num2str(e)))
for i=1:108
tbd
end
What is the function for checking each variable for > 1000 and saving it in the same variable?
Thank you in advance for your support and solution.I have 108 variables with same name but with consecutive numbering. Each of this variables I want to check if there is a value bigger 1000. When yes, then it will be exchanged through NaN.
I try to solve with eval and for-loop, but it doesn’t work.
I know the function which changes a value bigger 1000 through NaN.
Function: Temperatur_1(Temperatur_1>1000)=NaN;
Dynamics variabels are: Temperatur_1, Temperatur_2, etc.
eval-function: eval(strcat(‘Temperatur_’,num2str(e)))
for i=1:108
tbd
end
What is the function for checking each variable for > 1000 and saving it in the same variable?
Thank you in advance for your support and solution. I have 108 variables with same name but with consecutive numbering. Each of this variables I want to check if there is a value bigger 1000. When yes, then it will be exchanged through NaN.
I try to solve with eval and for-loop, but it doesn’t work.
I know the function which changes a value bigger 1000 through NaN.
Function: Temperatur_1(Temperatur_1>1000)=NaN;
Dynamics variabels are: Temperatur_1, Temperatur_2, etc.
eval-function: eval(strcat(‘Temperatur_’,num2str(e)))
for i=1:108
tbd
end
What is the function for checking each variable for > 1000 and saving it in the same variable?
Thank you in advance for your support and solution. eval, dynamic variable MATLAB Answers — New Questions

​

I am writing a program to seperate, plot even and odd parts of a discrete time signal. Following is my code:
I am getting error at line 3 & 4. "Array indices must be positive integers or logical values."
Since I am trying to flip and invert the signal x[n] , so ‘x(-n)’ is obvious to use but I am getting error. Kindly help.

n= -3:4;
x=[0 0 0 2 3 -1 2 -3];
xe= ( x(n)+x(-n) )/2;
xo= ( x(n)-x(-n) )/2;
subplot(311);stem(n,x);grid on; xlabel(‘n’);ylabel(‘Amplitude’);title(‘Original Signal’);
subplot(312);stem(n,xe);grid on; xlabel(‘n’);ylabel(‘Amplitude’);title(‘Even Signal’);
subplot(313);stem(n,xo);grid on; xlabel(‘n’);ylabel(‘Amplitude’);title(‘Odd Signal’);I am writing a program to seperate, plot even and odd parts of a discrete time signal. Following is my code:
I am getting error at line 3 & 4. "Array indices must be positive integers or logical values."
Since I am trying to flip and invert the signal x[n] , so ‘x(-n)’ is obvious to use but I am getting error. Kindly help.

n= -3:4;
x=[0 0 0 2 3 -1 2 -3];
xe= ( x(n)+x(-n) )/2;
xo= ( x(n)-x(-n) )/2;
subplot(311);stem(n,x);grid on; xlabel(‘n’);ylabel(‘Amplitude’);title(‘Original Signal’);
subplot(312);stem(n,xe);grid on; xlabel(‘n’);ylabel(‘Amplitude’);title(‘Even Signal’);
subplot(313);stem(n,xo);grid on; xlabel(‘n’);ylabel(‘Amplitude’);title(‘Odd Signal’); I am writing a program to seperate, plot even and odd parts of a discrete time signal. Following is my code:
I am getting error at line 3 & 4. "Array indices must be positive integers or logical values."
Since I am trying to flip and invert the signal x[n] , so ‘x(-n)’ is obvious to use but I am getting error. Kindly help.

n= -3:4;
x=[0 0 0 2 3 -1 2 -3];
xe= ( x(n)+x(-n) )/2;
xo= ( x(n)-x(-n) )/2;
subplot(311);stem(n,x);grid on; xlabel(‘n’);ylabel(‘Amplitude’);title(‘Original Signal’);
subplot(312);stem(n,xe);grid on; xlabel(‘n’);ylabel(‘Amplitude’);title(‘Even Signal’);
subplot(313);stem(n,xo);grid on; xlabel(‘n’);ylabel(‘Amplitude’);title(‘Odd Signal’); dicrete_time_signal MATLAB Answers — New Questions

​

As most of the sensor have some or other kind of time delay, in my model I have used a temperature sensor the signal provided by the sensor is going through a transfer function just to replicate its true nature (i.e thermal inertia). As the sensor data is being used in a feed back loop. The senor with a transfer function is causing eratic behaviour of my whole model. Because when the setpoint is high enough the sensor reading are all over the place. What should I do?As most of the sensor have some or other kind of time delay, in my model I have used a temperature sensor the signal provided by the sensor is going through a transfer function just to replicate its true nature (i.e thermal inertia). As the sensor data is being used in a feed back loop. The senor with a transfer function is causing eratic behaviour of my whole model. Because when the setpoint is high enough the sensor reading are all over the place. What should I do? As most of the sensor have some or other kind of time delay, in my model I have used a temperature sensor the signal provided by the sensor is going through a transfer function just to replicate its true nature (i.e thermal inertia). As the sensor data is being used in a feed back loop. The senor with a transfer function is causing eratic behaviour of my whole model. Because when the setpoint is high enough the sensor reading are all over the place. What should I do? transfer function, steady state error MATLAB Answers — New Questions

​

I want to perform the sensitivity analysis on the parameters of an ODE SIR model using normalized sensitivity function. I used the following code below. When it was run, I got this error: "undefined functionnor variable ‘p’, error in epidemic1 line8, F=ode45(@epidemic,ic,p)". How can I resolve it to get the plots?
My interest is to get the figure (3) i.e. The plot of nomalized sensitivity functions against time
function epidemic1()

clc

beta=0.4; gamma=0.04;
p1 = [beta gamma];
ic = [995;5;0];
F = ode45(@epidemic,ic,p);

sol1 = solve(F,0,80)

figure(1)
plot(sol1.Time,sol1.Solution,’-o’)
legend(‘S’,’I’,’R’)
title([‘SIR Populations over Time’,’$beta=0.4$, $gamma=0.04$’],’Interpreter’,’latex’)
xlabel(‘Time’,’Interpreter’,’latex’)
ylabel(‘Population’,’Interpreter’,’latex’)

Figure(2)
p2 = [0.2 0.1];
F.Parameters = p2;
F.Sensitivity = odeSensitivity;
sol2 = solve(F,0,80)

plot(sol2.Time,sol2.Solution,’-o’)
legend(‘S’,’I’,’R’)
title([‘SIR Populations over Time’,’beta=0.2$, $gamma=0.1$’],’Interpreter’,’latex’)
xlabel(‘Time’,’Interpreter’,’latex’)
ylabel(‘Population’,’Interpreter’,’latex’)

U11 = squeeze(sol2.Sensitivity(1,1,:))’.*(p2(1)./sol2.Solution(1,:));
U12 = squeeze(sol2.Sensitivity(1,2,:))’.*(p2(2)./sol2.Solution(1,:));
U21 = squeeze(sol2.Sensitivity(2,1,:))’.*(p2(1)./sol2.Solution(2,:));
U22 = squeeze(sol2.Sensitivity(2,2,:))’.*(p2(2)./sol2.Solution(2,:));
U31 = squeeze(sol2.Sensitivity(3,1,:))’.*(p2(1)./sol2.Solution(3,:));
U32 = squeeze(sol2.Sensitivity(3,2,:))’.*(p2(2)./sol2.Solution(3,:));

Figure(3)
t = tiledlayout(3,2);
title(t,[‘Parameter Sensitivity by Equation’,’$beta=0.2$, $gamma=0.1$’],’Interpreter’,’latex’)
xlabel(t,’Time’,’Interpreter’,’latex’)
ylabel(t,’% Change in Eqn’,’Interpreter’,’latex’)
nexttile
plot(sol2.Time,U11)
title(‘$beta$, Eqn. 1′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U12)
title(‘$gamma$, Eqn. 1′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U21)
title(‘$beta$, Eqn. 2′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U22)
title(‘$gamma$, Eqn. 2′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U31)
title(‘$beta$, Eqn. 3′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U32)
title(‘$gamma$, Eqn. 3′,’Interpreter’,’latex’)
ylim([-5 5])
function dydt = epidemic(t,y,p)
dydt = [0;0;0];
S = y(1);
I = y(2);
R = y(3);
N = S + I + R;

dSdt = -beta*(I*S)/N;
dIdt = beta*(I*S)/N – gamma*I;
dRdt = gamma*I;

end
endI want to perform the sensitivity analysis on the parameters of an ODE SIR model using normalized sensitivity function. I used the following code below. When it was run, I got this error: "undefined functionnor variable ‘p’, error in epidemic1 line8, F=ode45(@epidemic,ic,p)". How can I resolve it to get the plots?
My interest is to get the figure (3) i.e. The plot of nomalized sensitivity functions against time
function epidemic1()

clc

beta=0.4; gamma=0.04;
p1 = [beta gamma];
ic = [995;5;0];
F = ode45(@epidemic,ic,p);

sol1 = solve(F,0,80)

figure(1)
plot(sol1.Time,sol1.Solution,’-o’)
legend(‘S’,’I’,’R’)
title([‘SIR Populations over Time’,’$beta=0.4$, $gamma=0.04$’],’Interpreter’,’latex’)
xlabel(‘Time’,’Interpreter’,’latex’)
ylabel(‘Population’,’Interpreter’,’latex’)

Figure(2)
p2 = [0.2 0.1];
F.Parameters = p2;
F.Sensitivity = odeSensitivity;
sol2 = solve(F,0,80)

plot(sol2.Time,sol2.Solution,’-o’)
legend(‘S’,’I’,’R’)
title([‘SIR Populations over Time’,’beta=0.2$, $gamma=0.1$’],’Interpreter’,’latex’)
xlabel(‘Time’,’Interpreter’,’latex’)
ylabel(‘Population’,’Interpreter’,’latex’)

U11 = squeeze(sol2.Sensitivity(1,1,:))’.*(p2(1)./sol2.Solution(1,:));
U12 = squeeze(sol2.Sensitivity(1,2,:))’.*(p2(2)./sol2.Solution(1,:));
U21 = squeeze(sol2.Sensitivity(2,1,:))’.*(p2(1)./sol2.Solution(2,:));
U22 = squeeze(sol2.Sensitivity(2,2,:))’.*(p2(2)./sol2.Solution(2,:));
U31 = squeeze(sol2.Sensitivity(3,1,:))’.*(p2(1)./sol2.Solution(3,:));
U32 = squeeze(sol2.Sensitivity(3,2,:))’.*(p2(2)./sol2.Solution(3,:));

Figure(3)
t = tiledlayout(3,2);
title(t,[‘Parameter Sensitivity by Equation’,’$beta=0.2$, $gamma=0.1$’],’Interpreter’,’latex’)
xlabel(t,’Time’,’Interpreter’,’latex’)
ylabel(t,’% Change in Eqn’,’Interpreter’,’latex’)
nexttile
plot(sol2.Time,U11)
title(‘$beta$, Eqn. 1′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U12)
title(‘$gamma$, Eqn. 1′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U21)
title(‘$beta$, Eqn. 2′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U22)
title(‘$gamma$, Eqn. 2′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U31)
title(‘$beta$, Eqn. 3′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U32)
title(‘$gamma$, Eqn. 3′,’Interpreter’,’latex’)
ylim([-5 5])
function dydt = epidemic(t,y,p)
dydt = [0;0;0];
S = y(1);
I = y(2);
R = y(3);
N = S + I + R;

dSdt = -beta*(I*S)/N;
dIdt = beta*(I*S)/N – gamma*I;
dRdt = gamma*I;

end
end I want to perform the sensitivity analysis on the parameters of an ODE SIR model using normalized sensitivity function. I used the following code below. When it was run, I got this error: "undefined functionnor variable ‘p’, error in epidemic1 line8, F=ode45(@epidemic,ic,p)". How can I resolve it to get the plots?
My interest is to get the figure (3) i.e. The plot of nomalized sensitivity functions against time
function epidemic1()

clc

beta=0.4; gamma=0.04;
p1 = [beta gamma];
ic = [995;5;0];
F = ode45(@epidemic,ic,p);

sol1 = solve(F,0,80)

figure(1)
plot(sol1.Time,sol1.Solution,’-o’)
legend(‘S’,’I’,’R’)
title([‘SIR Populations over Time’,’$beta=0.4$, $gamma=0.04$’],’Interpreter’,’latex’)
xlabel(‘Time’,’Interpreter’,’latex’)
ylabel(‘Population’,’Interpreter’,’latex’)

Figure(2)
p2 = [0.2 0.1];
F.Parameters = p2;
F.Sensitivity = odeSensitivity;
sol2 = solve(F,0,80)

plot(sol2.Time,sol2.Solution,’-o’)
legend(‘S’,’I’,’R’)
title([‘SIR Populations over Time’,’beta=0.2$, $gamma=0.1$’],’Interpreter’,’latex’)
xlabel(‘Time’,’Interpreter’,’latex’)
ylabel(‘Population’,’Interpreter’,’latex’)

U11 = squeeze(sol2.Sensitivity(1,1,:))’.*(p2(1)./sol2.Solution(1,:));
U12 = squeeze(sol2.Sensitivity(1,2,:))’.*(p2(2)./sol2.Solution(1,:));
U21 = squeeze(sol2.Sensitivity(2,1,:))’.*(p2(1)./sol2.Solution(2,:));
U22 = squeeze(sol2.Sensitivity(2,2,:))’.*(p2(2)./sol2.Solution(2,:));
U31 = squeeze(sol2.Sensitivity(3,1,:))’.*(p2(1)./sol2.Solution(3,:));
U32 = squeeze(sol2.Sensitivity(3,2,:))’.*(p2(2)./sol2.Solution(3,:));

Figure(3)
t = tiledlayout(3,2);
title(t,[‘Parameter Sensitivity by Equation’,’$beta=0.2$, $gamma=0.1$’],’Interpreter’,’latex’)
xlabel(t,’Time’,’Interpreter’,’latex’)
ylabel(t,’% Change in Eqn’,’Interpreter’,’latex’)
nexttile
plot(sol2.Time,U11)
title(‘$beta$, Eqn. 1′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U12)
title(‘$gamma$, Eqn. 1′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U21)
title(‘$beta$, Eqn. 2′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U22)
title(‘$gamma$, Eqn. 2′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U31)
title(‘$beta$, Eqn. 3′,’Interpreter’,’latex’)
ylim([-5 5])
nexttile
plot(sol2.Time,U32)
title(‘$gamma$, Eqn. 3′,’Interpreter’,’latex’)
ylim([-5 5])
function dydt = epidemic(t,y,p)
dydt = [0;0;0];
S = y(1);
I = y(2);
R = y(3);
N = S + I + R;

dSdt = -beta*(I*S)/N;
dIdt = beta*(I*S)/N – gamma*I;
dRdt = gamma*I;

end
end sensisitivity analysis, ode, normalized sensitivity MATLAB Answers — New Questions

​

With a very large dataset – assume 1 million [x,y] pairs or more – how do I find the *index of an individual datapoint* on a MATLAB plot?

Example: I use plot(x,y) and the plot appears. The plot is complicated but there are artifacts I can locate by eye much easier than writing a detection script. I zoom into the plot and locate the desired data artifact. I need to know the exact location (index) of the artifact. I zoom that individual datapoints are apparent, place a cursor on the datapoint of interest, but the cursor does not report the exact x(index) of the datapoint. Other than manually scrubbing through the data to find the index, how do I find the index by using the plot?With a very large dataset – assume 1 million [x,y] pairs or more – how do I find the *index of an individual datapoint* on a MATLAB plot?

Example: I use plot(x,y) and the plot appears. The plot is complicated but there are artifacts I can locate by eye much easier than writing a detection script. I zoom into the plot and locate the desired data artifact. I need to know the exact location (index) of the artifact. I zoom that individual datapoints are apparent, place a cursor on the datapoint of interest, but the cursor does not report the exact x(index) of the datapoint. Other than manually scrubbing through the data to find the index, how do I find the index by using the plot? With a very large dataset – assume 1 million [x,y] pairs or more – how do I find the *index of an individual datapoint* on a MATLAB plot?

Example: I use plot(x,y) and the plot appears. The plot is complicated but there are artifacts I can locate by eye much easier than writing a detection script. I zoom into the plot and locate the desired data artifact. I need to know the exact location (index) of the artifact. I zoom that individual datapoints are apparent, place a cursor on the datapoint of interest, but the cursor does not report the exact x(index) of the datapoint. Other than manually scrubbing through the data to find the index, how do I find the index by using the plot? plot MATLAB Answers — New Questions

​

In my Simulink model, current value of a signal should be added to all previous value of the same signal. I’ve used integrator with initial value of zero to find it; however the result is different from what I find in Matlab through "sum" or "trapz" function.
The solver is continuous with fixed step size.
Any thought to resolve this issue. or any other block to use for this purpose instead of integrator?In my Simulink model, current value of a signal should be added to all previous value of the same signal. I’ve used integrator with initial value of zero to find it; however the result is different from what I find in Matlab through "sum" or "trapz" function.
The solver is continuous with fixed step size.
Any thought to resolve this issue. or any other block to use for this purpose instead of integrator? In my Simulink model, current value of a signal should be added to all previous value of the same signal. I’ve used integrator with initial value of zero to find it; however the result is different from what I find in Matlab through "sum" or "trapz" function.
The solver is continuous with fixed step size.
Any thought to resolve this issue. or any other block to use for this purpose instead of integrator? sum simulink MATLAB Answers — New Questions

​

Hi,

I am using the neural network fitting app in MATLAB 2024a version.

I used the code to develop the neural network by adding a hidden layer.

When I use this code, it shows the results for training, but I can’t get the results for the additional test set as shown in the neural network fitting app.

How can I get it? (e.g. the name of additional test input is "inputtest" and that of additional test output(target) is "outputtes")

I wrote my code below.

% Solve an Input-Output Fitting problem with a Neural Network
% Script generated by Neural Fitting app
% Created 09-Jun-2024 21:04:21
%
% This script assumes these variables are defined:
%
% bodyfatInputs – input data.
% bodyfatTargets – target data.

x = bodyfatInputs;
t = bodyfatTargets;

% Choose a Training Function
% For a list of all training functions type: help nntrain
% ‘trainlm’ is usually fastest.
% ‘trainbr’ takes longer but may be better for challenging problems.
% ‘trainscg’ uses less memory. Suitable in low memory situations.
trainFcn = ‘trainlm’; % Levenberg-Marquardt backpropagation.

% Create a Fitting Network
hiddenLayer1Size = 10;
hiddenLayer2Size = 10;
net = fitnet([hiddenLayer1Size, hiddenLayer2Size],trainFcn);

% Setup Division of Data for Training, Validation, Testing
net.divideParam.trainRatio = 70/100;
net.divideParam.valRatio = 15/100;
net.divideParam.testRatio = 15/100;

% Train the Network
[net,tr] = train(net,x,t);

% Test the Network
y = net(x);
e = gsubtract(t,y);
performance = perform(net,t,y)

% View the Network
view(net)

% Plots
% Uncomment these lines to enable various plots.
%figure, plotperform(tr)
%figure, plottrainstate(tr)
%figure, ploterrhist(e)
%figure, plotregression(t,y)
%figure, plotfit(net,x,t)Hi,

I am using the neural network fitting app in MATLAB 2024a version.

I used the code to develop the neural network by adding a hidden layer.

When I use this code, it shows the results for training, but I can’t get the results for the additional test set as shown in the neural network fitting app.

How can I get it? (e.g. the name of additional test input is "inputtest" and that of additional test output(target) is "outputtes")

I wrote my code below.

% Solve an Input-Output Fitting problem with a Neural Network
% Script generated by Neural Fitting app
% Created 09-Jun-2024 21:04:21
%
% This script assumes these variables are defined:
%
% bodyfatInputs – input data.
% bodyfatTargets – target data.

x = bodyfatInputs;
t = bodyfatTargets;

% Choose a Training Function
% For a list of all training functions type: help nntrain
% ‘trainlm’ is usually fastest.
% ‘trainbr’ takes longer but may be better for challenging problems.
% ‘trainscg’ uses less memory. Suitable in low memory situations.
trainFcn = ‘trainlm’; % Levenberg-Marquardt backpropagation.

% Create a Fitting Network
hiddenLayer1Size = 10;
hiddenLayer2Size = 10;
net = fitnet([hiddenLayer1Size, hiddenLayer2Size],trainFcn);

% Setup Division of Data for Training, Validation, Testing
net.divideParam.trainRatio = 70/100;
net.divideParam.valRatio = 15/100;
net.divideParam.testRatio = 15/100;

% Train the Network
[net,tr] = train(net,x,t);

% Test the Network
y = net(x);
e = gsubtract(t,y);
performance = perform(net,t,y)

% View the Network
view(net)

% Plots
% Uncomment these lines to enable various plots.
%figure, plotperform(tr)
%figure, plottrainstate(tr)
%figure, ploterrhist(e)
%figure, plotregression(t,y)
%figure, plotfit(net,x,t) Hi,

I am using the neural network fitting app in MATLAB 2024a version.

I used the code to develop the neural network by adding a hidden layer.

When I use this code, it shows the results for training, but I can’t get the results for the additional test set as shown in the neural network fitting app.

How can I get it? (e.g. the name of additional test input is "inputtest" and that of additional test output(target) is "outputtes")

I wrote my code below.

% Solve an Input-Output Fitting problem with a Neural Network
% Script generated by Neural Fitting app
% Created 09-Jun-2024 21:04:21
%
% This script assumes these variables are defined:
%
% bodyfatInputs – input data.
% bodyfatTargets – target data.

x = bodyfatInputs;
t = bodyfatTargets;

% Choose a Training Function
% For a list of all training functions type: help nntrain
% ‘trainlm’ is usually fastest.
% ‘trainbr’ takes longer but may be better for challenging problems.
% ‘trainscg’ uses less memory. Suitable in low memory situations.
trainFcn = ‘trainlm’; % Levenberg-Marquardt backpropagation.

% Create a Fitting Network
hiddenLayer1Size = 10;
hiddenLayer2Size = 10;
net = fitnet([hiddenLayer1Size, hiddenLayer2Size],trainFcn);

% Setup Division of Data for Training, Validation, Testing
net.divideParam.trainRatio = 70/100;
net.divideParam.valRatio = 15/100;
net.divideParam.testRatio = 15/100;

% Train the Network
[net,tr] = train(net,x,t);

% Test the Network
y = net(x);
e = gsubtract(t,y);
performance = perform(net,t,y)

% View the Network
view(net)

% Plots
% Uncomment these lines to enable various plots.
%figure, plotperform(tr)
%figure, plottrainstate(tr)
%figure, ploterrhist(e)
%figure, plotregression(t,y)
%figure, plotfit(net,x,t) neural network, neural networks fitting MATLAB Answers — New Questions

​

Hi,
I’m using the new R2024a "Segment Anything" model with good results.
However, it seems like the original Segment Anything Model by Meta has also the ‘everything’ option, where it somehow segments all the objects in the image with no user input.
Is there a way to use the Matlab version for a similar result? Just specifying a grid of ‘foreground points’ does not help, as the Matlab function considers them all to belong to the same object.
thanks in advanceHi,
I’m using the new R2024a "Segment Anything" model with good results.
However, it seems like the original Segment Anything Model by Meta has also the ‘everything’ option, where it somehow segments all the objects in the image with no user input.
Is there a way to use the Matlab version for a similar result? Just specifying a grid of ‘foreground points’ does not help, as the Matlab function considers them all to belong to the same object.
thanks in advance Hi,
I’m using the new R2024a "Segment Anything" model with good results.
However, it seems like the original Segment Anything Model by Meta has also the ‘everything’ option, where it somehow segments all the objects in the image with no user input.
Is there a way to use the Matlab version for a similar result? Just specifying a grid of ‘foreground points’ does not help, as the Matlab function considers them all to belong to the same object.
thanks in advance image segmentation, deep learning, r2024a MATLAB Answers — New Questions

​

Hello everyone,
So far I have dealt with the RRT* algorithm that received a map (a binary matrix that I turned into a map) and the initial and final position of the robot.
Now I want to use the Potential Field algorithm and put in it a start and end position and in addition the robot’s environment (the map)
How can i do it? does not succeed.
This is the code for the RRT*
clc; clear;close
% Load the original 25×25 matrix
original_matrix = dlmread("C:UsersuserDesktopmap.txt", ‘t’);
% Scale and increase dimensions to 50×50
scaled_matrix = imresize(original_matrix, [50, 50], ‘nearest’);
a=occupancyMap(scaled_matrix);
inflatedmap=copy(a);
inflate(inflatedmap,10^-1)
startPose = getRandomPose(inflatedmap);
goalPose = getRandomPose(inflatedmap);
[path,treeData] = codegenRRTStar1(inflatedmap,startPose,goalPose);
I want all the black positions the positions where there is a 1 in the matrix to be the obstacles

Attached is a link to the implementation of the desired algorithm in Matlab
https://www.mathworks.com/matlabcentral/fileexchange/126455-artificial-potential-field-path-planning-algorithm

Many thanks to all the helpers

*The positions are absolute in space and do not specify a row/column in the matrix.Hello everyone,
So far I have dealt with the RRT* algorithm that received a map (a binary matrix that I turned into a map) and the initial and final position of the robot.
Now I want to use the Potential Field algorithm and put in it a start and end position and in addition the robot’s environment (the map)
How can i do it? does not succeed.
This is the code for the RRT*
clc; clear;close
% Load the original 25×25 matrix
original_matrix = dlmread("C:UsersuserDesktopmap.txt", ‘t’);
% Scale and increase dimensions to 50×50
scaled_matrix = imresize(original_matrix, [50, 50], ‘nearest’);
a=occupancyMap(scaled_matrix);
inflatedmap=copy(a);
inflate(inflatedmap,10^-1)
startPose = getRandomPose(inflatedmap);
goalPose = getRandomPose(inflatedmap);
[path,treeData] = codegenRRTStar1(inflatedmap,startPose,goalPose);
I want all the black positions the positions where there is a 1 in the matrix to be the obstacles

Attached is a link to the implementation of the desired algorithm in Matlab
https://www.mathworks.com/matlabcentral/fileexchange/126455-artificial-potential-field-path-planning-algorithm

Many thanks to all the helpers

*The positions are absolute in space and do not specify a row/column in the matrix. Hello everyone,
So far I have dealt with the RRT* algorithm that received a map (a binary matrix that I turned into a map) and the initial and final position of the robot.
Now I want to use the Potential Field algorithm and put in it a start and end position and in addition the robot’s environment (the map)
How can i do it? does not succeed.
This is the code for the RRT*
clc; clear;close
% Load the original 25×25 matrix
original_matrix = dlmread("C:UsersuserDesktopmap.txt", ‘t’);
% Scale and increase dimensions to 50×50
scaled_matrix = imresize(original_matrix, [50, 50], ‘nearest’);
a=occupancyMap(scaled_matrix);
inflatedmap=copy(a);
inflate(inflatedmap,10^-1)
startPose = getRandomPose(inflatedmap);
goalPose = getRandomPose(inflatedmap);
[path,treeData] = codegenRRTStar1(inflatedmap,startPose,goalPose);
I want all the black positions the positions where there is a 1 in the matrix to be the obstacles

Attached is a link to the implementation of the desired algorithm in Matlab
https://www.mathworks.com/matlabcentral/fileexchange/126455-artificial-potential-field-path-planning-algorithm

Many thanks to all the helpers

*The positions are absolute in space and do not specify a row/column in the matrix. maze, map, occupancymap, potential field algorithm MATLAB Answers — New Questions

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function t_a = SecOrdReaction (t0,tf,y,N)
%Решение обыкновенных дифференциальных уравнений для мономолекулярных реакций второго порядка
% t – время, сек; у – количество вещества А, М
% t_a = [0 600]- интервал времени; y = 0,25 – начальное условие;
% k = 0.5 – константа скорости реакции.
% N-число точек для отображения на графике
% Для запуска функции необходимо в командном окне набрать: >> t_a = SecOrdReaction (0,600,0.25,5)
close all %закрыть все ранее построенные графические окна
t = [t0, tf];
[t, y] = ode45(‘ODEfirst’, t0, y);
plot(t, y)
xlabel(‘Время, c’);
ylabel(‘Концентрация вещества A, M’)
title( ‘Решение кинетического уравнения для мономолекулярной реакции второго порядка’)
grid on
t_n = linpack(t0, tf, n);% при решении будет использовано точек интервала времени [ts, tf]
y_n = interp1 (t, y, t_n, ‘spline’);% n значений y точек для отображения
t_A = [t_n’, y_n’];
hold on
plot(t_n,y_n,’*r’)
function dy = ODEfirst(t, y)
dy=-0.5*y2;
[t, y] = ode45 (‘ODEfirst’, [0: 100: 600], 0.25) % t= 0 … 600 сек, шаг=100; y = 0.25
t_a = SecOrdReaction (0,600,0.25,5)function t_a = SecOrdReaction (t0,tf,y,N)
%Решение обыкновенных дифференциальных уравнений для мономолекулярных реакций второго порядка
% t – время, сек; у – количество вещества А, М
% t_a = [0 600]- интервал времени; y = 0,25 – начальное условие;
% k = 0.5 – константа скорости реакции.
% N-число точек для отображения на графике
% Для запуска функции необходимо в командном окне набрать: >> t_a = SecOrdReaction (0,600,0.25,5)
close all %закрыть все ранее построенные графические окна
t = [t0, tf];
[t, y] = ode45(‘ODEfirst’, t0, y);
plot(t, y)
xlabel(‘Время, c’);
ylabel(‘Концентрация вещества A, M’)
title( ‘Решение кинетического уравнения для мономолекулярной реакции второго порядка’)
grid on
t_n = linpack(t0, tf, n);% при решении будет использовано точек интервала времени [ts, tf]
y_n = interp1 (t, y, t_n, ‘spline’);% n значений y точек для отображения
t_A = [t_n’, y_n’];
hold on
plot(t_n,y_n,’*r’)
function dy = ODEfirst(t, y)
dy=-0.5*y2;
[t, y] = ode45 (‘ODEfirst’, [0: 100: 600], 0.25) % t= 0 … 600 сек, шаг=100; y = 0.25
t_a = SecOrdReaction (0,600,0.25,5) function t_a = SecOrdReaction (t0,tf,y,N)
%Решение обыкновенных дифференциальных уравнений для мономолекулярных реакций второго порядка
% t – время, сек; у – количество вещества А, М
% t_a = [0 600]- интервал времени; y = 0,25 – начальное условие;
% k = 0.5 – константа скорости реакции.
% N-число точек для отображения на графике
% Для запуска функции необходимо в командном окне набрать: >> t_a = SecOrdReaction (0,600,0.25,5)
close all %закрыть все ранее построенные графические окна
t = [t0, tf];
[t, y] = ode45(‘ODEfirst’, t0, y);
plot(t, y)
xlabel(‘Время, c’);
ylabel(‘Концентрация вещества A, M’)
title( ‘Решение кинетического уравнения для мономолекулярной реакции второго порядка’)
grid on
t_n = linpack(t0, tf, n);% при решении будет использовано точек интервала времени [ts, tf]
y_n = interp1 (t, y, t_n, ‘spline’);% n значений y точек для отображения
t_A = [t_n’, y_n’];
hold on
plot(t_n,y_n,’*r’)
function dy = ODEfirst(t, y)
dy=-0.5*y2;
[t, y] = ode45 (‘ODEfirst’, [0: 100: 600], 0.25) % t= 0 … 600 сек, шаг=100; y = 0.25
t_a = SecOrdReaction (0,600,0.25,5) matlab MATLAB Answers — New Questions

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